This article takes an in-depth look at how Class C power amplifiers work and compares them to Class A and Class B amplifiers.
Over the years, several power amplifier topologies have been developed to meet the needs of different applications. Some of these amplifier categories, including the one we will discuss, are based on their on-angle (θc) Class A and Class B amplifiers have been discussed in the previous articles in this series. In a Class A amplifier, the transistor is always on, so the conduction angle is 360 degrees. In a Class B amplifier, the transistor is turned on only half the time of the signal cycle, and the conduction angle is 180 degrees.
We learned that by reducing the turn-on angle, we could increase the efficiency from 50% for Class A to 78.5% for Class B. But what happens to the efficiency if the conduction angle is further reduced?
Power amplifiers with a conduction angle of less than 180 degrees are known as Class C amplifiers. In this article, we will analyze the working principle of Class C amplifiers in detail and discuss the effects of reduced conduction angles on various performance parameters of power amplifiers. Finally, we will compare the performance of Class A, Class B, and Class C amplifiers through a classic comprehensive analysis.
Current and voltage waveforms in Class C amplifiers
In Class C amplifiers, the transistor takes less than half the time to turn on during the input cycle. Stimulated by a narrow pulse, the transistor produces a brief pulse of current at the output. The orange curve in Figure 1 shows the collector current for one cycle of a Class C amplifier at the on-angle θc.
Figure 1.A current waveform with a conduction angle of class θc
As you can see from the diagram, when the transistor is active, the output current is part of the sine wave, and when the transistor is cut off, the output current is zero. The full sine wave (blue curve in the figure) has a negative offset IQ and an amplitude IRF. The DC offset IQ here is similar to the bias current in a linear amplifier, but is negative in a Class C amplifier. Therefore, the output current can be described by the following expression:
By changing the IQ, we can also generate waveforms for Class A and Class B amplifiers. For example, when IQ=0, the conduction angle is 180 degrees (Class B). Therefore, we can use the above waveforms to examine the performance of all three amplifier classes (Class A, Class B, and Class C).
Figure 2 compares the transistor current and output voltage waveforms of near-perfect Class A, B, and C amplifiers.
Figure 2.Current (a) and voltage (b) waveforms for Class A, B, and C amplifiers
Circuit schematic diagram of a Class C amplifier
Figure 3 shows the basic circuit schematic of a Class C amplifier. The high-Q resonator at the amplifier output is marked in green.
Figure 3. Basic schematic diagram of a Class C amplifier
Depending on the static bias point we have chosen for the transistor, the above schematic can also be used to build a Class A amplifier or a single transistor Class B amplifier. Class A amplifiers are the most linear of the three types and may use resonant circuits with relatively low Q factors.
At the other end of the linear spectrum, the Class C amplifier generates a series of short current pulses at the output. The high-Q resonant circuit short-circuits output current harmonics and reduce out-of-band emissions caused by unavoidable nonlinearities. Note that a high-Q resonant circuit necessarily implies narrow-band operation.
Waveform analysis
When analyzing Class C amplifiers, we make the following assumptions:
1. Output voltage waveform: The output voltage can be approximated as a sinusoidal waveform. This requires an ideal resonant circuit to short-circuit all the higher harmonics of the output current.
2. Collector current waveform: The collector current waveform is a part of the sine wave. In fact, this only holds at low frequencies.
Although these assumptions are not necessarily entirely accurate in practice, they allow us to simplify circuit analysis. With that in mind, let's examine the performance of Class C amplifiers.
Due to the action of the high-Q resonator, the power actually transferred to the load is located at the fundamental frequency. Therefore, in order to find the output power, we need to analyze the frequency component of the output current waveform. By changing the time origin of the waveform, we can simplify this analysis. Figure 4 shows the collector current for one cycle in a Class C amplifier with respect to the vertical axis.
Figure 4: Collector current for one cycle in a Class C amplifier
The above waveform can be described as a cosine function as:
Using the Fourier series, we can express the output current in the form of its constituent frequency components:
where an denotes the Fourier coefficient of the nth harmonic. In order to find the efficiency and output power of a Class C amplifier, we only need to know the average value of the current waveform ((a0) and the fundamental component (a1). Without expanding all the mathematical calculations in detail, the result is shown by the following formula:
where φ is equal to half of the conduction angle
Figure 5 illustrates the average and fundamental components as a function of the conduction angle.
Figure 5.Average and basic components as a function of conduction angle.
The graph shows the results after normalizing the a0 and a1 coefficients to IM (or equivalent, assuming IM is unit 1). We'll come back to those results later. First, let's calculate the efficiency of a Class C amplifier.
Efficiency of Class C amplifiers
Assuming that a high-Q resonator eliminates the higher harmonic components, the AC output voltage can be calculated by Equation 5 as:
where RL is the load resistance.
Therefore, the average power delivered to the load is:
To calculate the power provided by a power supply, we multiply the average of the current drawn from the power supply by the supply voltage. The average value of the current is a0, therefore:
From Equations 7 and 8, we can calculate the efficiency:
With the appropriate load resistance, the fundamental component of the current produces the maximum allowable voltage swing. The amplitude of the maximum voltage swing is VCC. Therefore, as can be seen from Equation 6, efficiency is maximized when the following conditions are met:
Since the higher harmonic components are eliminated by the high-Q resonator, the AC output voltage and power can be calculated using the Fourier series to analyze the frequency component of the output current. Through a series of mathematical calculations, we can obtain the formula for the maximum efficiency of a Class C amplifier and plot it as a curve in Figure 6.
Figure 6.
Next, we will further illustrate the application of these formulas with an example.
Example: Selecting a maximum current specification for a Class C amplifier
We know that transistors are limited in terms of the maximum voltage and current levels they can handle, as well as the maximum power they can withstand without damage. Assuming the maximum transistor current of a Class C amplifier, the amplifier delivers 25W into a 50Ω load with a maximum efficiency of 85%. Ignore the saturation effect of the transistor and assume a supply voltage (VCC) of 12V.
The fundamental component of the output current (a1) determines the power transferred to the load. From Equation 7, we have:
In this example, we are working at maximum efficiency. Therefore, we know that RLa1 = VCC (Equation 10). Since VCC=12V, we have RLa1 = VCC=12V. Substituting the value of 12V into Equation 13, we can get the amplitude of the fundamental component:
From Figure 7, which reproduces the plot of maximum efficiency versus conduction angle, we observe that ηmax=85% corresponds to θc=147∘.
Figure 7. At a conduction angle of θc=147∘, the maximum efficiency reaches 85%.
At a conduction angle of θc=147∘, the normalized value of A1 is 0.45 (Figure 8).
Figure 8. At a conduction angle of 147∘, the normalized fundamental component is 0.45.
In other words, we have:
The transistor should be able to handle a maximum current of 9.27A. The maximum voltage experienced by the transistor is 24V, which is twice the supply voltage (2VCC=2×12V=24V).
Compare Class A, Class B, and Class C operations
By changing the conduction angle, we can observe the effect on the following performance parameters:
- Power supply
- Output power
- Maximum efficiency
Decreasing the conduction angle reduces the average power drawn from the power supply, but it also reduces the output power. However, the efficiency increases as the conduction angle decreases, although the output power also decreases accordingly.
How does the power of the power supply change with thetac?
Figure 5 shows that the DC component of the output current decreases monotonically as the conduction angle decreases from 360 degrees (Class A operation) to 180 degrees (Class B) and then to 0 degrees.
This makes sense if we consider the current waveform in Figure 4. A smaller on-angle means a smaller area where the current is non-zero, which also corresponds to a smaller average. As a result, decreasing the conduction angle reduces the average value and the DC power drawn from the power supply.
How does the load power change with θc?
The fundamental component in Figure 5 exhibits more interesting behavior. At 360 degrees, the value of the fundamental component is 0.5. As the conduction angle decreases from 360 degrees to 180 degrees, the fundamental component increases slightly.
However, at 180 degrees, the fundamental component value is again 0.5. This means that both Class A and Class B amplifiers produce the same maximum output power at the same transistor specifications and supply voltages.
What about the C-type work area? If we substitute Equation 10 into Equation 7, we can observe that the maximum output power is proportional to A1. As can be seen in Figure 5, as θc approaches zero, a1 also approaches zero. As a result, the output power of the Class C amplifier is also reduced to zero. This is a significant disadvantage of Class C operations.
How does the efficiency of a Class C amplifier change with theta?
In the Class C operating region, both the mean and fundamental components decrease as the conduction angle decreases. From Equation 11, we know that the maximum efficiency (ηmax) is proportional to the ratio of the fundamental component to the mean. At the same time, it can be confirmed by visual inspection Figure 6 that ηmax increases as the conduction angle decreases. When the conduction angle is close to zero, the efficiency is close to 100%.
Although this result may seem exciting at first glance, keep in mind that the output power of a Class C amplifier is much lower than that of a Class A or Class B amplifier for the same input power. For example, even if a Class C amplifier is as efficient as 95%, if it produces only half the power of the same Class A amplifier, then the practical use of this configuration is limited.
Disadvantages of Class C amplifiers
There are a few other limitations to the Class C configuration:
1. For a given output power, the transistors used in a Class C amplifier must handle a greater current than the transistors in a Class A or B amplifier. This gets worse as we reduce the conduction angle to improve efficiency.
2. A larger maximum current means we need a larger device, resulting in a lower matching bandwidth.
3. Class C configurations are more non-linear than Class A or Class B amplifiers.
4. The Class C stage requires a transistor with a higher breakdown voltage.
5. Compared to Class A or Class B amplifiers, the Class C stage requires a higher Q resonant circuit to suppress the harmonic components.
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